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#25
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![]() Quote:
Here's an example that shows you can't do the calc like you did it above. Say we have 3 horses, except I'm going to identify them by the cards J, Q, and K. After shuffling, you will pick them 1 at a time for win and place. What is the chance a J will “finish” either 1st or 2nd? Common sense says that the J has 2/3 chance of finishing 1st or 2nd in a 3-card "race". Same for Q or K. What’s the chance that J and Q BOTH finish in the top 2? By your methodology, it would be 2/3 * 2/3 = 44.4%. But it’s easy to show that 44.4% is wrong. There are only 3 possible pairs that can hold the top 2 places: JQ, JK, and QK. Each of those 3 pairs must have the same chance, namely 1/3 chance. So the chance of J AND Q finishing in the top 2 spots is 33.3%, not 44.4%. Similarly, your ".8 x .5 = .4" is an over-estimate of the combined chance of both horses finishing in the top 3. So, given your 80%/50% assumptions for Curlin and Hard Spun, the bet is even better for you than you thought. --Dunbar
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Curlin and Hard Spun finish 1,2 in the 2007 BC Classic, demonstrating how competing in all three Triple Crown races ruins a horse for the rest of the year...see avatar photo from REUTERS/Lucas Jackson |